\(\int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx\) [153]

Optimal result

Integrand size = 26, antiderivative size = 123 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c}+\frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {\left (b^2-4 a c\right )^2 d^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{3/2}} \] Output:

-1/16*(-4*a*c+b^2)*d^2*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*d^2*(2*c*x+b)^3 
*(c*x^2+b*x+a)^(1/2)/c-1/32*(-4*a*c+b^2)^2*d^2*arctanh(1/2*(2*c*x+b)/c^(1/ 
2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.82 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {d^2 \left (\sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (b^2+8 b c x+4 c \left (a+2 c x^2\right )\right )-\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )\right )}{16 c^{3/2}} \] Input:

Integrate[(b*d + 2*c*d*x)^2*Sqrt[a + b*x + c*x^2],x]
 

Output:

(d^2*(Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(b^2 + 8*b*c*x + 4*c*(a + 
2*c*x^2)) - (b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b 
+ c*x)])]))/(16*c^(3/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1109, 27, 1116, 1092, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {a+b x+c x^2} (b d+2 c d x)^2 \, dx\)

\(\Big \downarrow \) 1109

\(\displaystyle \frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {\left (b^2-4 a c\right ) \int \frac {d^2 (b+2 c x)^2}{\sqrt {c x^2+b x+a}}dx}{16 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {d^2 \left (b^2-4 a c\right ) \int \frac {(b+2 c x)^2}{\sqrt {c x^2+b x+a}}dx}{16 c}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {d^2 \left (b^2-4 a c\right ) \left (\frac {1}{2} \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx+(b+2 c x) \sqrt {a+b x+c x^2}\right )}{16 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {d^2 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}+(b+2 c x) \sqrt {a+b x+c x^2}\right )}{16 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{8 c}-\frac {d^2 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+(b+2 c x) \sqrt {a+b x+c x^2}\right )}{16 c}\)

Input:

Int[(b*d + 2*c*d*x)^2*Sqrt[a + b*x + c*x^2],x]
 

Output:

(d^2*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/(8*c) - ((b^2 - 4*a*c)*d^2*((b + 
 2*c*x)*Sqrt[a + b*x + c*x^2] + ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt 
[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c])))/(16*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1109

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1)))   Int[(d + e*x)^m*(a + b*x 
 + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b* 
e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1 
)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && IntegerQ[2*p]
 

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+8 a \,c^{2} x +10 b^{2} c x +4 a b c +b^{3}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}}{16 c}-\frac {\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{2}}{32 c^{\frac {3}{2}}}\)	\(114\)
default	\(d^{2} \left (b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+4 c^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+4 b c \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\right )\)	\(354\)

Input:

int((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/c*(16*c^3*x^3+24*b*c^2*x^2+8*a*c^2*x+10*b^2*c*x+4*a*b*c+b^3)*(c*x^2+b 
*x+a)^(1/2)*d^2-1/32*(16*a^2*c^2-8*a*b^2*c+b^4)/c^(3/2)*ln((1/2*b+c*x)/c^( 
1/2)+(c*x^2+b*x+a)^(1/2))*d^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.51 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\left [\frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} d^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} d^{2} x^{3} + 24 \, b c^{3} d^{2} x^{2} + 2 \, {\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{2} x + {\left (b^{3} c + 4 \, a b c^{2}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{64 \, c^{2}}, \frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (16 \, c^{4} d^{2} x^{3} + 24 \, b c^{3} d^{2} x^{2} + 2 \, {\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{2} x + {\left (b^{3} c + 4 \, a b c^{2}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{32 \, c^{2}}\right ] \] Input:

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/64*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*d^2*log(-8*c^2*x^2 - 8*b*c*x 
 - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(16*c^4* 
d^2*x^3 + 24*b*c^3*d^2*x^2 + 2*(5*b^2*c^2 + 4*a*c^3)*d^2*x + (b^3*c + 4*a* 
b*c^2)*d^2)*sqrt(c*x^2 + b*x + a))/c^2, 1/32*((b^4 - 8*a*b^2*c + 16*a^2*c^ 
2)*sqrt(-c)*d^2*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2 
*x^2 + b*c*x + a*c)) + 2*(16*c^4*d^2*x^3 + 24*b*c^3*d^2*x^2 + 2*(5*b^2*c^2 
 + 4*a*c^3)*d^2*x + (b^3*c + 4*a*b*c^2)*d^2)*sqrt(c*x^2 + b*x + a))/c^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (110) = 220\).

Time = 0.57 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.08 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \cdot \left (\frac {3 b c d^{2} x^{2}}{2} + c^{2} d^{2} x^{3} + \frac {x \left (a c^{2} d^{2} + \frac {5 b^{2} c d^{2}}{4}\right )}{2 c} + \frac {a b c d^{2} + b^{3} d^{2} - \frac {3 b \left (a c^{2} d^{2} + \frac {5 b^{2} c d^{2}}{4}\right )}{4 c}}{c}\right ) + \left (a b^{2} d^{2} - \frac {a \left (a c^{2} d^{2} + \frac {5 b^{2} c d^{2}}{4}\right )}{2 c} - \frac {b \left (a b c d^{2} + b^{3} d^{2} - \frac {3 b \left (a c^{2} d^{2} + \frac {5 b^{2} c d^{2}}{4}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \cdot \left (\frac {4 c^{2} d^{2} \left (a + b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 8 a c^{2} d^{2} + 4 b^{2} c d^{2}\right )}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \cdot \left (4 a^{2} c^{2} d^{2} - 4 a b^{2} c d^{2} + b^{4} d^{2}\right )}{3 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {4 \sqrt {a} c^{2} d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:

integrate((2*c*d*x+b*d)**2*(c*x**2+b*x+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x + c*x**2)*(3*b*c*d**2*x**2/2 + c**2*d**2*x**3 + x* 
(a*c**2*d**2 + 5*b**2*c*d**2/4)/(2*c) + (a*b*c*d**2 + b**3*d**2 - 3*b*(a*c 
**2*d**2 + 5*b**2*c*d**2/4)/(4*c))/c) + (a*b**2*d**2 - a*(a*c**2*d**2 + 5* 
b**2*c*d**2/4)/(2*c) - b*(a*b*c*d**2 + b**3*d**2 - 3*b*(a*c**2*d**2 + 5*b* 
*2*c*d**2/4)/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x 
**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) 
+ x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 0)), (2*(4*c**2*d**2*(a + b*x 
)**(7/2)/(7*b**2) + (a + b*x)**(5/2)*(-8*a*c**2*d**2 + 4*b**2*c*d**2)/(5*b 
**2) + (a + b*x)**(3/2)*(4*a**2*c**2*d**2 - 4*a*b**2*c*d**2 + b**4*d**2)/( 
3*b**2))/b, Ne(b, 0)), (4*sqrt(a)*c**2*d**2*x**3/3, True))
 

Maxima [F(-2)]

Exception generated. \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {1}{16} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c^{2} d^{2} x + 3 \, b c d^{2}\right )} x + \frac {5 \, b^{2} c^{3} d^{2} + 4 \, a c^{4} d^{2}}{c^{3}}\right )} x + \frac {b^{3} c^{2} d^{2} + 4 \, a b c^{3} d^{2}}{c^{3}}\right )} + \frac {{\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{32 \, c^{\frac {3}{2}}} \] Input:

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/16*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c^2*d^2*x + 3*b*c*d^2)*x + (5*b^2*c^3* 
d^2 + 4*a*c^4*d^2)/c^3)*x + (b^3*c^2*d^2 + 4*a*b*c^3*d^2)/c^3) + 1/32*(b^4 
*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2*d^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + 
 b*x + a))*sqrt(c) + b))/c^(3/2)
 

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.72 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=b^2\,d^2\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-a\,c\,d^2\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )-\frac {5\,b\,c\,d^2\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{2}+c\,d^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}+\frac {b\,d^2\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{4\,c^{3/2}}+\frac {b\,d^2\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{6\,c}+\frac {b^2\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}} \] Input:

int((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(1/2),x)
 

Output:

b^2*d^2*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) - a*c*d^2*((x/2 + b/(4*c)) 
*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1 
/2))*(a*c - b^2/4))/(2*c^(3/2))) - (5*b*c*d^2*((log((b + 2*c*x)/c^(1/2) + 
2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^ 
2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/2 + c*d^2*x*(a + 
 b*x + c*x^2)^(3/2) + (b*d^2*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2) 
^(1/2))*(b^3 - 4*a*b*c))/(4*c^(3/2)) + (b*d^2*(8*c*(a + c*x^2) - 3*b^2 + 2 
*b*c*x)*(a + b*x + c*x^2)^(1/2))/(6*c) + (b^2*d^2*log((b/2 + c*x)/c^(1/2) 
+ (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.07 \[ \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx=\frac {d^{2} \left (8 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2}+16 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x +2 \sqrt {c \,x^{2}+b x +a}\, b^{3} c +20 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x +48 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{2}+32 \sqrt {c \,x^{2}+b x +a}\, c^{4} x^{3}-16 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}+8 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c -\sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}\right )}{32 c^{2}} \] Input:

int((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(1/2),x)
 

Output:

(d**2*(8*sqrt(a + b*x + c*x**2)*a*b*c**2 + 16*sqrt(a + b*x + c*x**2)*a*c** 
3*x + 2*sqrt(a + b*x + c*x**2)*b**3*c + 20*sqrt(a + b*x + c*x**2)*b**2*c** 
2*x + 48*sqrt(a + b*x + c*x**2)*b*c**3*x**2 + 32*sqrt(a + b*x + c*x**2)*c* 
*4*x**3 - 16*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sq 
rt(4*a*c - b**2))*a**2*c**2 + 8*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x* 
*2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c - sqrt(c)*log((2*sqrt(c)*sqr 
t(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4))/(32*c**2)